جیسے ایک جبل گرد کو جبل گرد پر چڑھنا ہے - کیونکہ وہ یہاں ہے، اسی طرح ایک اچھا ریاضی کا طالب علم نئی مواد کو سیکھتا ہے کیونکہ وہ یہاں ہے۔ - جیمز بی۔ برسٹول

7.1 تعارف

تفاضلی حساب کا مرکزی مفہرست تفاضل کا مفہرست ہے۔ تفاضل کے اصل حوافز ایسے مسائل کے لیے تھے جو تفاعلیں کی گراف کے ٹینجنٹ لائنوں کی تعریف کرنے اور ان لائنوں کی میل کی معلوم کرنے کے لیے تھے۔ تکاملی حساب کو ایسے مسائل کے حوافز سے حاصل کیا گیا ہے جو تفاعلیں کی گراف کے تفاعل کے تحت حدود میں شکل دینے والے علاقے کی تعریف کرنے اور معلوم کرنے کے لیے تھے۔

اگر ایک تفاعل $f$ ایک دائرہ $I$ میں قابل تفاضل ہے، یعنی اس کا تفاضل $f$ $I$ کے ہر پونچہ میں موجود ہے، تو ایک طبیعی سوال پیش آتا ہے کہ $f^{\prime}$ ہر پونچے پر I، ہم تفاعل کا معیار کر سکتے ہیں؟ ان تفاعلات کو جو کہ اس تفاعل کے تفاضل کے بدلے ہو سکتے ہیں، اسے تفاعل کے بدلے کے (یا اصل) تفاعل کہا جاتا ہے۔ اس کے علاوہ، جو صیغہ جو https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/images/ncertbook/math/m12/integrals/ncert_m12_ch07_figure_g_w_leibnitz.png"

جی۔ ایس۔ لیبنیز (1646 - 1716)

ان تمام تفاعل کے بدلے کو ایک تفاعل کے بدلے کہا جاتا ہے اور ان تفاعل کے بدلے کی تلاش کا عمل تکامل کہلاتا ہے۔ ایسی قسم کی مسائل بہت سی عملی حالات میں آتی ہیں۔ مثال کے طور پر، اگر ہمیں کسی جسم کی لحظانی تیزی کا معلوم ہو تو، ایک طبیعی سوال پیش آتا ہے، یعنی، ہم کسی بھی لحظہ میں جسم کی حالت کا معیار کر سکتے ہیں؟ تکامل کے عمل میں شامل ہونے والی بہت سی عملی اور نظریہ حالات ہیں۔ تکاملی حساب کا تعمیر اینے مسائل کا حل کرنے کے امتحان کے نتیجے میں ہوا:

(أ) ایک تفاعل کی تلاش کرنے کا مسئلہ جب اس کا تفاضل دیا گیا ہو،

(ب) ایک تفاعل کی گراف کے تحت حدود میں شکل دینے والی علاقے کی تلاش کرنے کا مسئلہ ایسی شرائط کے تحت۔

ان دو مسائل سے دو قسم کے تکاملات، مثلاً غیر معین اور معین تکاملات، حاصل ہوتے ہیں، جو مل کر تکاملی حساب کی تشکیل دیتے ہیں۔

غیر معین تکامل اور معین تکامل کے درمیان ایک جوڑا، جس کو تکاملی حساب کا بنیادی مسل کہا جاتا ہے، معین تکامل کو علم اور ماہرین کے لیے ایک عملی اوزار کے طور پر بناتا ہے۔ معین تکامل کو بہت سی جذبات سے منفرد مسائل کو حل کرنے کے لیے بھی استعمال کیا جاتا ہے، جیسے انقلاب، مالیات اور تقسیماتیات۔

اس فصل میں، ہم غیر معین اور معین تکامل اور ان کے بنیادی خصوصیات کو جاننے کے لیے محدود رہیں گے، شامل کر کے تکامل کے کچھ طریقے۔

7.2 تفاضل کے عکس عمل کے طور پر تکامل

تکامل تفاضل کے عکس عمل ہے۔ ایک تفاعل کو تفاضل کرنے کے بجائے، ہمیں ایک تفاعل کا تفاضل دیا جاتا ہے اور اس کا اصل تفاعل، یعنی اصل تفاعل کی تلاش کرنے کے لیے کہا جاتا ہے۔ ایسا عمل تکامل یا تفاضل کے عکس کہلاتا ہے۔ دیکھیں:

$$ \begin{equation*} \frac{d}{d x}\left(\frac{x^{3}}{3}\right)=x^{2} \tag{2} \end{equation*} $$

$$ \frac{d}{d x}(\sin x+C)=\cos x, \frac{d}{d x}(\frac{x^{3}}{3}+C)=x^{2} \text{ and } \frac{d}{d x}(e^{x}+C)=e^{x} $$

$$ \frac{d}{d x} \int f(x) d x=f(x) $$

$$ \frac{d}{d x} F(x)=f(x) $$

$$ \text{ }\qquad \int f(x) d x=F(x)+C $$

$$ f^{\prime}(x)=\frac{d}{d x} f(x) $$

$$\frac{d}{d x} \int f(x) d x=\frac{d}{d x} \int g(x) d x$$

$$ \int(f(x)+g(x)) d x=\int f(x) d x+\int g(x) d x . $$

$$ \begin{aligned} & \int[k_1 f_1(x)+k_2 f_2(x)+\ldots+k_n f_n(x)] d x \\ & =k_1 \int f_1(x) d x+k_2 \int f_2(x) d x+\ldots+k_n \int f_n(x) d x . \end{aligned} $$

$$ \begin{aligned} \int \frac{x^{3}-1}{x^{2}} & d x=\int x d x-\int x^{-2} d x \quad(\text{ by Property } V) \\ = & (\frac{x^{1+1}}{1+1}+C_1)-(\frac{x^{-2+1}}{-2+1}+C_2) ; C_1, C_2 \text{ are constants of integration } \\ & =\frac{x^{2}}{2}+C_1-\frac{x^{-1}}{-1}-C_2=\frac{x^{2}}{2}+\frac{1}{x}+C_1-C_2 \\ & =\frac{x^{2}}{2}+\frac{1}{x}+C \text{, where } C=C_1-C_2 \text{ is another constant of integration. } \end{aligned} $$

$$ \begin{aligned} \int(x^{\frac{2}{3}}+1) d x & =\int x^{\frac{2}{3}} d x+\int d x \\ & =\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+x+C=\frac{3}{5} x^{\frac{5}{3}}+x+C \end{aligned} $$

$$ \begin{aligned} & =\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+C \\ & =\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+C \end{aligned} $$

$$ \begin{aligned} \int(\sin x+\cos x) d x & =\int \sin x d x+\int \cos x d x \\ & =-\cos x+\sin x+C \end{aligned} $$

$$ \begin{aligned} \int(cosec x(cosec x+\cot x) d x & =\int cosec^{2} x d x+\int cosec x \cot x d x \\ & =-\cot x-cosec x+C \end{aligned} $$

$$ \begin{aligned} \int \frac{1-\sin x}{\cos ^{2} x} d x & =\int \frac{1}{\cos ^{2} x} d x-\int \frac{\sin x}{\cos ^{2} x} d x \\ & =\int \sec ^{2} x d x-\int \tan x \sec x d x \\ & =\tan x-\sec x+C \end{aligned} $$

$$ \frac{d}{d x}(x^{4}-6 x)=4 x^{3}-6 $$

$$ F(x)=x^{4}-6 x+C \text{, where } C \text{ is constant. } $$

$$ \begin{aligned} F(0) & =3, \text{ which gives } \\ 3 & =0-6 \times 0+C \text{ or } C=3 \end{aligned} $$

$$ \int y^{4} d y=\frac{y^{4+1}}{4+1}+C=\frac{1}{5} y^{5}+C $$

$$ \begin{equation*} \frac{d}{d x}\left(\frac{x^{3}}{3}\right)=x^{2} \tag{2} \end{equation*} $$

$$ \frac{d}{d x}(\sin x+C)=\cos x, \frac{d}{d x}(\frac{x^{3}}{3}+C)=x^{2} \text{ and } \frac{d}{d x}(e^{x}+C)=e^{x} $$

$$ \frac{d}{d x} \int f(x) d x=f(x) $$

$$ \frac{d}{d x} F(x)=f(x) $$

$$ \text{ }\qquad \int f(x) d x=F(x)+C $$

$$ f^{\prime}(x)=\frac{d}{d x} f(x) $$

$$\frac{d}{d x} \int f(x) d x=\frac{d}{d x} \int g(x) d x$$

$$ \int(f(x)+g(x)) d x=\int f(x) d x+\int g(x) d x . $$

$$ \begin{aligned} & \int[k_1 f_1(x)+k_2 f_2(x)+\ldots+k_n f_n(x)] d x \\ & =k_1 \int f_1(x) d x+k_2 \int f_2(x) d x+\ldots+k_n \int f_n(x) d x . \end{aligned} $$

$$ \begin{aligned} \int \frac{x^{3}-1}{x^{2}} & d x=\int x d x-\int x^{-2} d x \quad(\text{ by Property } V) \\ = & (\frac{x^{1+1}}{1+1}+C_1)-(\frac{x^{-2+1}}{-2+1}+C_2) ; C_1, C_2 \text{ are constants of integration } \\ & =\frac{x^{2}}{2}+C_1-\frac{x^{-1}}{-1}-C_2=\frac{x^{2}}{2}+\frac{1}{x}+C_1-C_2 \\ & =\frac{x^{2}}{2}+\frac{1}{x}+C \text{, where } C=C_1-C_2 \text{ is another constant of integration. } \end{aligned} $$

$$ \begin{aligned} \int(x^{\frac{2}{3}}+1) d x & =\int x^{\frac{2}{3}} d x+\int d x \\ & =\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}+x+C=\frac{3}{5} x^{\frac{5}{3}}+x+C \end{aligned} $$

$$ \begin{aligned} & =\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+2 e^{x}-\log |x|+C \\ & =\frac{2}{5} x^{\frac{5}{2}}+2 e^{x}-\log |x|+C \end{aligned} $$

$$ \begin{aligned} \int(\sin x+\cos x) d x & =\int \sin x d x+\int \cos x d x \\ & =-\cos x+\sin x+C \end{aligned} $$

$$ \begin{aligned} \int(cosec x(cosec x+\cot x) d x & =\int cosec^{2} x d x+\int cosec x \cot x d x \\ & =-\cot x-cosec x+C \end{aligned} $$

$$ \begin{aligned} \int \frac{1-\sin x}{\cos ^{2} x} d x & =\int \frac{1}{\cos ^{2} x} d x-\int \frac{\sin x}{\cos ^{2} x} d x \\ & =\int \sec ^{2} x d x-\int \tan x \sec x d x \\ & =\tan x-\sec x+C \end{aligned} $$

$$ \frac{d}{d x}(x^{4}-6 x)=4 x^{3}-6 $$

$$ F(x)=x^{4}-6 x+C \text{, where } C \text{ is constant. } $$

$$ \begin{aligned} F(0) & =3, \text{ which gives } \\ 3 & =0-6 \times 0+C \text{ or } C=3 \end{aligned} $$

$$ \int y^{4} d y=\frac{y^{4+1}}{4+1}+C=\frac{1}{5} y^{5}+C $$

$$ I=\int f(x) d x $$

$$ d x=g^{\prime}(t) d t $$

$$ I=\int f(x) d x=\int f(g(t)) g^{\prime}(t) d t $$

$$\int 2 x \sin (x^{2}+1) d x=\int \sin t d t=-\cos t+C=-\cos (x^{2}+1)+C$$

$$ \begin{aligned} & =\frac{2}{5} \tan ^{5} t+\mathrm{C}(\text { क्योंकि } u=\tan t) \\ & =\frac{2}{5} \tan ^{5} \sqrt{x}+\mathrm{C}(\text { क्योंकि } t=\sqrt{x}) \end{aligned} $$

$$ \int \cot x d x=\int \frac{d t}{t}=\log |t|+C=\log |\sin x|+C $$

$$ \begin{aligned} \int \sin ^{3} x \cos ^{2} x d x & =\int \sin ^{2} x \cos ^{2} x(\sin x) d x \\ & =\int(1-\cos ^{2} x) \cos ^{2} x(\sin x) d x \end{aligned} $$

$$ \begin{aligned} & =-\int(t^{2}-t^{4}) d t=-(\frac{t^{3}}{3}-\frac{t^{5}}{5})+C \\ & =-\frac{1}{3} \cos ^{3} x+\frac{1}{5} \cos ^{5} x+C \end{aligned} $$

$$ \begin{aligned} \int \frac{\sin x}{\sin (x+a)} d x & =\int \frac{\sin (t-a)}{\sin t} d t \\ & =\int \frac{\sin t \cos a-\cos t \sin a}{\sin t} d t \\ & =\cos a \int d t-\sin a \int \cot t d t \\ & =(\cos a) t-(\sin a)[\log |\sin t|+C_1] \\ & =(\cos a)(x+a)-(\sin a)[\log |\sin (x+a)|+C_1] \\ & =x \cos a+a \cos a-(\sin a) \log |\sin (x+a)|-C_1 \sin a \end{aligned} $$

$$ \begin{aligned} & =\frac{1}{2} \int \frac{(\cos x+\sin x+\cos x-\sin x) d x}{\cos x+\sin x} \\ & =\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \\ & =\frac{x}{2}+\frac{C_1}{2}+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \end{aligned} $$

$$ \begin{aligned} \int \frac{d x}{1+\tan x} & =\frac{x}{2}+\frac{C_1}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{C_2}{2} \\ & =\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{C_1}{2}+\frac{C_2}{2} \\ & =\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+C,(C=\frac{C_1}{2}+\frac{C_2}{2}) \end{aligned} $$

$$ \cos ^{2} x=\frac{1+\cos 2 x}{2} $$

$$ =\frac{x}{2}+\frac{1}{4} \sin 2 x+C $$

$$ \begin{aligned} $\int \sin 2 x \cos 3 x d x=\frac{1}{2}[\int \sin 5 x d x-\int \sin x d x]$\\ & =\frac{1}{2}[-\frac{1}{5} \cos 5 x+\cos x]+C \\ & =-\frac{1}{10} \cos 5 x+\frac{1}{2} \cos x+C \end{aligned} $$

$$ =-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x+C $$

$$ =-\cos x+\frac{1}{3} \cos ^{3} x+C $$

$$ =\frac{1}{2 a}[\frac{(x+a)-(x-a)}{(x-a)(x+a)}]=\frac{1}{2 a}[\frac{1}{x-a}-\frac{1}{x+a}] $$

$$ \begin{aligned} & =\frac{1}{2 a}[\log |(x-a)|-\log |(x+a)|]+C \\ & =\frac{1}{2 a} \log |\frac{x-a}{x+a}|+C \end{aligned} $$

$$ \begin{aligned} & =\frac{1}{2 a}[-\log |a-x|+\log |a+x|]+C \\ & =\frac{1}{2 a} \log |\frac{a+x}{a-x}|+C \end{aligned} $$

$$ =\frac{1}{a} \int d \theta=\frac{1}{a} \theta+C=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C $$

$$ \begin{aligned} \int \frac{d x}{\sqrt{x^{2}+a^{2}}} & =\int \frac{a \sec ^{2} \theta d \theta}{\sqrt{a^{2} \tan ^{2} \theta+a^{2}}} \\ & \text{ Therefore, } \qquad =\int \sec \theta d \theta=\log |(\sec \theta+\tan \theta)|+C_1 \end{aligned} $$

$$ \begin{aligned} & =\log |\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}+1}|+C_1 \\ & =\log |x+\sqrt{x^{2}+a^{2}}|-\log |a|+C_1 \\ & =\log |x+\sqrt{x^{2}+a^{2}}|+C, \text{ where } C=C_1-\log |a| \end{aligned} $$

$$ =\frac{1}{2} \tan ^{-1} \frac{x-3}{2}+C $$

$$ \begin{aligned} 3 x^{2}+13 x-10 & =3(x^{2}+\frac{13 x}{3}-\frac{10}{3}) \\ & =3[(x+\frac{13}{6})^{2}-(\frac{17}{6})^{2}] \text{ (completing the square) } \end{aligned} $$

$$ \begin{aligned} & =\frac{1}{3 \times 2 \times \frac{17}{6}} \log |\frac{t-\frac{17}{6}}{t+\frac{17}{6}}|+C_1 \quad \text{ [by } 7.4 \text{ (i)}] \\ & =\frac{1}{17} \log |\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}|+C_1 \\ & =\frac{1}{17} \log |\frac{6 x-4}{6 x+30}|+C_1 \\ & =\frac{1}{17} \log |\frac{3 x-2}{x+5}|+C_1+\frac{1}{17} \log \frac{1}{3} \\ & =\frac{1}{17} \log |\frac{3 x-2}{x+5}|+C, \text{ where } C=C_1+\frac{1}{17} \log \frac{1}{3} \end{aligned} $$

$$ \begin{align*} & =\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+\mathrm{C} \tag{4}\ & =\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{x^{2}-\frac{2 x}{5}}\right|+\mathrm{C} \end{align*} $$

$$ =\frac{1}{4} I_1+\frac{1}{2} I_2 \quad \text{ (say) } $$

$$ C=\frac{C_1}{4}+\frac{C_2}{2} $$

$$ \begin{equation*} =2 \sqrt{5-4 x-x^{2}}+\mathrm{C} _{1} \tag{2} \end{equation*} $$

$$ \frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2} $$

$$ 1=A(x+2)+B(x+1) . $$

$$ \begin{matrix} A+B=0 \\ \text{ And } \qquad 2 A+B=1 \end{matrix} $$

$$ \begin{aligned} \frac{1}{(x+1)(x+2)} & =\frac{1}{x+1}+\frac{-1}{x+2} \\ \int \frac{d x}{(x+1)(x+2)} & =\int \frac{d x}{x+1}-\int \frac{d x}{x+2} \\ & \text{ Therefore, } \qquad =\log |x+1|-\log |x+2|+C \\ & =\log |\frac{x+1}{x+2}|+C \end{aligned} $$

$$ \begin{aligned} \text{ let } \qquad \frac{x^{2}+1}{x^{2}-5 x+6} & =1+\frac{5 x-5}{x^{2}-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)}\\ \text{ So that }\frac{5 x-5}{(x-2)(x-3)} & =\frac{A}{x-2}+\frac{B}{x-3} \end{aligned} $$

$$ \frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3} \text{ writes } $$

$$ =\mathrm{A}\left(x^{2}+4 x+3\right)+\mathrm{B}(x+3)+\mathrm{C}\left(x^{2}+2 x+1\right) $$

$$ \frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)} $$

$$ \begin{aligned} \int \frac{3 x-2}{(x+1)^{2}(x+3)} & =\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3} \\ & =\frac{11}{4} \log |x+1|+\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+\mathrm{C} \\ & =\frac{11}{4} \log \left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+\mathrm{C} \end{aligned} $$

$$ \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}=\frac{y}{(y+1)(y+4)} $$

$$ \frac{y}{(y+1)(y+4)}=\frac{A}{y+1}+\frac{B}{y+4} $$

$$ y=A(y+4)+B(y+1) $$

$$ A=-\frac{1}{3} \quad \text{ and } \quad B=\frac{4}{3} $$

$$ \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}=-\frac{1}{3(x^{2}+1)}+\frac{4}{3(x^{2}+4)} $$

$$ \begin{aligned} \int \frac{x^{2} d x}{(x^{2}+1)(x^{2}+4)} & =-\frac{1}{3} \int \frac{d x}{x^{2}+1}+\frac{4}{3} \int \frac{d x}{x^{2}+4} \\ & =-\frac{1}{3} \tan ^{-1} x+\frac{4}{3} \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+C \\ & =-\frac{1}{3} \tan ^{-1} x+\frac{2}{3} \tan ^{-1} \frac{x}{2}+C \end{aligned} $$

$$ d y=\cos \phi d \phi $$

$$ \begin{aligned} & =\int \frac{3 y-2}{y^{2}-4 y+4} d y \\ & =\int \frac{3 y-2}{(y-2)^{2}}=I \text{ (say) } \end{aligned} $$

$$ \frac{3 y-2}{(y-2)^{2}}=\frac{A}{y-2}+\frac{B}{(y-2)^{2}} $$

$$ 3 y-2=A(y-2)+B $$

$$ \begin{aligned} I & =\int[\frac{3}{y-2}+\frac{4}{(y-2)^{2}}] d y=3 \int \frac{d y}{y-2}+4 \int \frac{d y}{(y-2)^{2}} \\ & =3 \log |y-2|+4(-\frac{1}{y-2})+C \\ & =3 \log |\sin \phi-2|+\frac{4}{2-\sin \phi}+C \\ & =3 \log (2-\sin \phi)+\frac{4}{2-\sin \phi}+C \text{ since, } 2-\sin \phi \text{ is always positive } \end{aligned} $$

$$ \frac{x^{2}+x+1}{(x^{2}+1)(x+2)}=\frac{A}{x+2}+\frac{B x+C}{(x^{2}+1)} $$

$$ x^{2}+x+1=A(x^{2}+1)+(B x+C)(x+2) $$

$$ =\frac{3}{5} \log |x+2|+\frac{1}{5} \log |x^{2}+1|+\frac{1}{5} \tan ^{-1} x+C $$

$$ \frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x} $$

$$ u v=\int u \frac{d v}{d x} d x+\int v \frac{d u}{d x} d x $$

$$ \int u \frac{d v}{d x} d x=u v-\int v \frac{d u}{d x} d x \tag{1} $$

$$ \begin{aligned} u & =f(x) \text{ and } \frac{d v}{d x}=g(x) . \text{ Then } \\ \frac{d u}{d x} & =f^{\prime}(x) \text{ and } v=\int g(x) d x \end{aligned} $$

$$ \int f(x) g(x) d x=f(x) \int g(x) d x-\int\left[\int g(x) d x f^{\prime}(x)\right] d x $$

$$ \int f(x) g(x) d x=f(x) \int g(x) d x-\int\left[f^{\prime}(x) \int g(x) d x\right] d x $$

$$ I_1=e^{x} \sin x-\int e^{x} \sin x d x $$

$$ A(x)=\int_a^{x} f(x) d x \tag{1} $$

$$ I=F(3)-F(2)=\frac{27}{3}-\frac{8}{3}=\frac{19}{3} $$

$$\text{ or } \qquad =\frac{1}{8}\left[u^{4}\right]=\frac{1}{8} \sin ^{4} 2 t=\mathrm{F}(t) \text { मान लीजिए } $$

$$ I=F(\frac{\pi}{4})-F(0)=\frac{1}{8}[\sin ^{4} \frac{\pi}{2}-\sin ^{4} 0]=\frac{1}{8} $$

$$ \begin{aligned} \int 5 x^{4} \sqrt{x^{5}+1} d x & =\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}=\frac{2}{3}\left(x^{5}+1\right)^{\frac{3}{2}} \\ \int _{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x & =\frac{2}{3}\left[\left(x^{5}+1\right)^{\frac{3}{2}}\right] _{-1}^{1} \\ & =\frac{2}{3}\left[\left(1^{5}+1\right)^{\frac{3}{2}}-\left((-1)^{5}+1\right)^{\frac{3}{2}}\right] \\ & =\frac{2}{3}\left[2^{\frac{3}{2}}-0^{\frac{3}{2}}\right]=\frac{2}{3}(2 \sqrt{2})=\frac{4 \sqrt{2}}{3} \end{aligned} $$

$$ \begin{align*} & \int _{a}^{b} f(x) d x=\mathrm{F}(b)-\mathrm{F}(a) \tag{1}\\ & \int _{a}^{c} f(x) d x=\mathrm{F}(c)-\mathrm{F}(a) \tag{2} \end{align*} $$

$$ \begin{equation*} \int _{c}^{b} f(x) d x=\mathrm{F}(b)-\mathrm{F}(c) \tag{2} \end{equation*} $$

$$ \int_a^{b} f(x) d x=-\int_b^{a} f(a+b-t) d t $$

$$ \begin{aligned} & =\int_a^{b} f(a+b-t) d t \quad(\text{ by } P_1) \\ & =\int_a^{b} f(a+b-x) d x \text{ by } P_0 \end{aligned} $$

$$\int_0^{2 a} f(x) d x=\int_0^{a} f(x) d x+\int_a^{2 a} f(x) d x$$

$$ \begin{aligned} \int _{a}^{2 a} f(x) d x & =-\int _{a}^{0} f(2 a-t) d t \\ & =\int _{0}^{a} f(2 a-t) d t=\int _{0}^{a} f(2 a-x) d x \text { is received } \end{aligned} $$

$$ \begin{equation*} \int _{0}^{2 a} f(x) d x=\int _{0}^{a} f(x) d x+\int _{0}^{a} f(2 a-x) d x \tag{1} \end{equation*} $$

$$ f(2 a-x)=f(x) \text{, then (1) becomes } $$

$$ \int_0^{2 a} f(x) d x=\int_0^{a} f(x) d x+\int_0^{a} f(x) d x=2 \int_0^{a} f(x) d x, $$

$$ f(2 a-x)=-f(x) \text{, then (1) becomes } $$

$$ \int_0^{2 a} f(x) d x=\int_0^{a} f(x) d x-\int_0^{a} f(x) d x=0 $$

$$ \int _{-a}^{a} f(x) d x=\int _{-a}^{0} f(x) d x+\int_0^{a} f(x) d x \text{. Then } $$

$$ \begin{aligned} d t & =-d x . \text{ When } x=-a, t=a \text{ and when } \\ x & =0, t=0 . \text{ Also } x=-t . \end{aligned} $$

$$ \int _{-a}^{a} f(x) d x=\int_0^{a} f(x) d x+\int_0^{a} f(x) d x=2 \int_0^{a} f(x) d x $$

$$ \int _{-a}^{a} f(x) d x=-\int_0^{a} f(x) d x+\int_0^{a} f(x) d x=0 $$

$$ \begin{align*} \int _{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^{2} x d x & =2 \int _{0}^{\frac{\pi}{4}} \sin ^{2} x d x \tag{7}\\ & =2 \int _{0}^{\frac{\pi}{4}} \frac{(1-\cos 2 x)}{2} d x=\int _{0}^{\frac{\pi}{4}}(1-\cos 2 x) d x \\ & =\left[x-\frac{1}{2} \sin 2 x\right] _{0}^{\frac{\pi}{4}}=\left(\frac{\pi}{4}-\frac{1}{2} \sin \frac{\pi}{2}\right)-0=\frac{\pi}{4}-\frac{1}{2} \end{align*} $$

$$ \begin{aligned} I & =\int_0^{\pi} \frac{(\pi-x) \sin (\pi-x) d x}{1+\cos ^{2}(\pi-x)} \\ & =\int_0^{\pi} \frac{(\pi-x) \sin x d x}{1+\cos ^{2} x}=\pi \int_0^{\pi} \frac{\sin x d x}{1+\cos ^{2} x}-I \end{aligned} $$

$$\text{ or } \qquad 2 I=\pi \int_0^{\pi} \frac{\sin x d x}{1+\cos ^{2} x} $$

$$\text{ or } \qquad I=\frac{\pi}{2} \int_0^{\pi} \frac{\sin x d x}{1+\cos ^{2} x} $$

$$ \begin{aligned} I & =\frac{-\pi}{2} \int_1^{-1} \frac{d t}{1+t^{2}}=\frac{\pi}{2} \int _{-1}^{1} \frac{d t}{1+t^{2}} \\ & =\pi \int_0^{1} \frac{d t}{1+t^{2}}(\text{ by } P_7, \text{ since } \frac{1}{1+t^{2}} \text{ is even function }) \\ & =\pi[\tan ^{-1} t]_0^{1}=\pi[\tan ^{-1} 1-\tan ^{-1} 0]=\pi[\frac{\pi}{4}-0]=\frac{\pi^{2}}{4} \end{aligned} $$

$$ \begin{align*} I & =\int _{0}^{\frac{\pi}{2}} \frac{\sin ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x \tag{1}\\ & =\int _{0}^{\frac{\pi}{2}} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x \tag{2} \end{align*} $$

$$ 2 I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x=\int_0^{\frac{\pi}{2}} d x=[x]_0^{\frac{\pi}{2}}=\frac{\pi}{2} $$

$$ I=\frac{\pi}{4} $$

$$ \begin{align*} \mathrm{I} & =\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\sqrt{\cos \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\sin \left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} \tag{1}\\ & =\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \tag{2} \end{align*} $$

$$ I=\int_0^{\frac{\pi}{2}} \log \sin (\frac{\pi}{2}-x) d x=\int_0^{\frac{\pi}{2}} \log \cos x d x $$

$$ \begin{aligned} 2 I= & \frac{1}{2} \int_0^{\pi/2} \log \sin t d t-\frac{\pi}{2} \log 2 \\ & =\frac{2}{2} \int_0^{\frac{\pi}{2}} \log \sin t d t-\frac{\pi}{2} \log 2 \quad[\text{ by } P_6 \text{ as } \sin (\pi-t)=\sin t) \\ & \text{ Therefore } \qquad =\int_0^{\frac{\pi}{2}} \log \sin x d x-\frac{\pi}{2} \log 2 \text{ (by changing variable) } t \text{ to } x) \\ & =I-\frac{\pi}{2} \log 2 \\ & \int _0^{\frac{\pi}{2}} \log \sin x d x=\frac{-\pi}{2} \log 2 . \end{aligned} $$

$$ \begin{aligned} \frac{x^{4}}{(x-1)(x^{2}+1)} & =(x+1)+\frac{1}{x^{3}-x^{2}+x-1} \\ & =(x+1)+\frac{1}{(x-1)(x^{2}+1)} \end{aligned} $$

$$ \frac{1}{(x-1)(x^{2}+1)}=\frac{A}{(x-1)}+\frac{B x+C}{(x^{2}+1)} \tag{2} $$

$$ \begin{aligned} \text{ So } \qquad 1 & =A(x^{2}+1)+(B x+C)(x-1) \\ & =(A+B) x^{2}+(C-B) x+A-C \end{aligned} $$

$$ \begin{equation*} \frac{1}{(x-1)\left(x^{2}+1\right)}=\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{\left(x^{2}+1\right)}-\frac{1}{2\left(x^{2}+1\right)} \tag{3} \end{equation*} $$

$$ \frac{x^{4}}{(x-1)(x^{2}+x+1)}=(x+1)+\frac{1}{2(x-1)}-\frac{1}{2} \frac{x}{(x^{2}+1)}-\frac{1}{2(x^{2}+1)} $$

$$ \int \frac{x^{4}}{(x-1)(x^{2}+x+1)} d x=\frac{x^{2}}{2}+x+\frac{1}{2} \log |x-1|-\frac{1}{4} \log (x^{2}+1)-\frac{1}{2} \tan ^{-1} x+C $$

$$ =\int \log (\log x) d x+\int \frac{1}{(\log x)^{2}} d x $$

$$ \begin{align*} \mathrm{I} & =x \log (\log x)-\int \frac{1}{x \log x} x d x+\int \frac{d x}{(\log x)^{2}} \\ & =x \log (\log x)-\int \frac{d x}{\log x}+\int \frac{d x}{(\log x)^{2}} \tag{1} \end{align*} $$

$$ \begin{equation*} \int \frac{d x}{\log x}=\left[\frac{x}{\log x}-\int x\left\{-\frac{1}{(\log x)^{2}}\left(\frac{1}{x}\right)\right\} d x\right] \tag{2} \end{equation*} $$

$$ \begin{aligned} \mathrm{I} & =x \log (\log x)-\frac{x}{\log x}-\int \frac{d x}{(\log x)^{2}}+\int \frac{d x}{(\log x)^{2}} \\ & =x \log (\log x)-\frac{x}{\log x}+\mathrm{C} \end{aligned} $$

$$\text{ or } \qquad d x=\frac{2 t d t}{1+t^{4}} $$

$$ \begin{aligned} \text{ Then } \qquad I & =\int t(1+\frac{1}{t^{2}}) \frac{2 t}{(1+t^{4})} d t \\ & =2 \int \frac{(t^{2}+1)}{t^{4}+1} d t=2 \int \frac{(1+\frac{1}{t^{2}}) d t}{(t^{2}+\frac{1}{t^{2}})}=2 \int \frac{(1+\frac{1}{t^{2}}) d t}{(t-\frac{1}{t})^{2}+2} \end{aligned} $$

$$t-\frac{1}{t}=y$, so that $(1+\frac{1}{t^{2}}) d t=d y$$

$$ \begin{aligned} \text{ Then } \qquad I & =2 \int \frac{d y}{y^{2}+(\sqrt{2})^{2}}=\sqrt{2} \tan ^{-1} \frac{y}{\sqrt{2}}+C=\sqrt{2} \tan ^{-1} \frac{(t-\frac{1}{t})}{\sqrt{2}}+C \\ & =\sqrt{2} \tan ^{-1}(\frac{t^{2}-1}{\sqrt{2} t})+C=\sqrt{2} \tan ^{-1}(\frac{\tan x-1}{\sqrt{2 \tan x}})+C \end{aligned} $$

$$\text{ Put } \qquad \cos ^{2}(2 x)=t \text { so that } 4 \sin 2 x \cos 2 x d x=-d t $$

$$\text{ Therefore } \qquad\quad I=-\frac{1}{4} \int \frac{d t}{\sqrt{9-t^{2}}}=-\frac{1}{4} \sin ^{-1}(\frac{t}{3})+C=-\frac{1}{4} \sin ^{-1}[\frac{1}{3} \cos ^{2} 2 x]+C$$

$$ \begin{aligned} \text{ Therefore } \qquad \int _ {-1}^{\frac{3}{2}}|x \sin \pi x| d x & =\int _ {-1}^{1} x \sin \pi x d x+\int _ 1^{\frac{3}{2}}-x \sin \pi x d x \\ & =\int _ {-1}^{1} x \sin \pi x d x-\int _ 1^{\frac{3}{2}} x \sin \pi x d x \end{aligned} $$

$$ \begin{aligned} & =\pi \int_0^{\pi} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}-\int_0^{\pi} \frac{x d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} \\ & =\pi \int_0^{\pi} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}-I \end{aligned} $$

$$\text{ Thus } \qquad\quad 2 I=\pi \int_0^{\pi} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}$$

$$ \begin{aligned} \text{ or } \qquad I & =\frac{\pi}{2} \int_ 0^{\pi} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}=\frac{\pi}{2} \cdot 2 \int_ 0^{\frac{\pi}{2}} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}(using P_ {6}) \\ & =\pi[\int_ 0^{\frac{\pi}{4}} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}+\int_ {\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}] \\ & =\pi[\int_0^{\frac{\pi}{4}} \frac{\sec ^{2} x d x}{a^{2}+b^{2} \tan ^{2} x}+\int_ {\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{cosec^{2} x d x}{a^{2} \cot ^{2} x+b^{2}}] \\ & =\pi[\int_ 0^{1} \frac{d t}{a^{2}+b^{2} t^{2}}-\int_ 1^{0} \frac{d u}{a^{2} u^{2}+b^{2}}]^{(p u t \tan x=tand \cot x=u)} \\ & =\frac{\pi}{a b}[\tan ^{-1} \frac{b t}{a}]_ 0^{1}-\frac{\pi}{a b}[\tan ^{-1} \frac{a u}{b}]_ 1^{0}=\frac{\pi}{a b}[\tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{a}{b}]=\frac{\pi^{2}}{2 a b} \end{aligned} $$

$$ \begin{aligned} & \int\left[k _{1} f _{1}(x)+k _{2} f _{2}(x)+\ldots+k _{n} f _{n}(x)\right] d x \\ & =k _{1} \int f _{1}(x) d x+k _{2} \int f _{2}(x) d x+\ldots+k _{n} \int f _{n}(x) d x \end{aligned} $$